java - 带输入的 PowerMockito mockStatic

Question

我需要模拟一个具有输入参数的静态方法。对于类中的其他函数，必须调用原始方法，对于我尝试模拟的函数，只必须执行模拟的 stub 。我尝试了以下代码，但它没有按预期工作。

public class Foo {
    public static String static1() {
        System.out.println("static1 called");
        return "1";
    }

    public static String static2() {
        System.out.println("static2 called");
        return "2";
    }

    public static String staticInput(int i) {
        System.out.println("staticInput called");
        return "static " + i;
    }
}

@RunWith(PowerMockRunner.class)
@PrepareForTest({Foo.class })
public class TestMockito {

    @Test
    public void test() throws Exception {

        PowerMockito.mockStatic(Foo.class, Mockito.CALLS_REAL_METHODS);
        PowerMockito.doReturn("dummy").when(Foo.class ,"static1");

        PowerMockito.when(Foo.staticInput(anyInt())).thenAnswer(invocation -> {
            System.out.println((int)invocation.getArgument(0));
            return "staticInput mock";
        });


        //        PowerMockito.doAnswer(new Answer() {
        //            @Override
        //            public Object answer(InvocationOnMock invocation) throws Throwable {
        //                int i = (int) invocation.getArguments()[0];
        //                System.out.println(i);
        //                return i;
        //            }
        //
        //        }).when(Foo.staticInput(anyInt()));

        System.out.println(Foo.static1());
        System.out.println(Foo.static2());
        System.out.println(Foo.staticInput(7));
    }
}

我得到以下输出:

staticInput called  
dummy 
static2 called  
2 
staticInput called  
static 7

Answer 1

我想出的最干净的代码是明确标记应该转发到其实际实现的方法。

PowerMockito.mockStatic(Foo.class);
PowerMockito.doReturn("dummy").when(Foo.class, "static1");
PowerMockito.when(Foo.static2()).thenCallRealMethod();
PowerMockito.when(Foo.staticInput(anyInt())).thenAnswer(invocation -> {
    System.out.println((int)invocation.getArgument(0));
    return "staticInput mock";
});

输出(符合我的期望):

dummy
static2 called
2
7
staticInput mock

奇怪的是，我的原始代码输出与您的输出不同(并显示带有输入参数的静态方法被模拟。):

staticInput called
dummy
static2 called
2
7
staticInput mock

我仍然相信我提出的版本更好:设置模拟时不会调用真正的静态方法，不幸的是，您的代码会发生这种情况。